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indica

indica

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indica
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  • Lagrangian is f−λg = 4x²+y²+z²−λ(2x+3y+z−11) For optimality we need ( 8x, 2y, 2z ) = λ(2,3,1) → x=λ/4, y=3λ/2, z=λ/2 Sub in constraint equation gives λ(1/2+9/2+1/2) = 11 → λ=2 Hence only solution is ( 1/2, 3, 1 ) with f=11 At fe…
    in minima and maxima problem? Comment by indica June 2021
  • Lagrangian is f−λg = 4x²+y²+z²−λ(2x+3y+z−11) For optimality we need ( 8x, 2y, 2z ) = λ(2,3,1) → x=λ/4, y=3λ/2, z=λ/2 Sub in constraint equation gives λ(1/2+9/2+1/2) = 11 → λ=2 Hence only solution is ( 1/2, 3, 1 ) with f=11 At fe…
    in minima and maxima problem? Comment by indica June 2021
  • Your 2nd step is wrong because k√p = sign(k)*√(k²p) So a*√(a²) = sign(a)*sqrt(a⁴)
    in does | a | x a = a^2? Comment by indica May 2021
  • Your 2nd step is wrong because k√p = sign(k)*√(k²p) So a*√(a²) = sign(a)*sqrt(a⁴)
    in does | a | x a = a^2? Comment by indica May 2021
  • Your 2nd step is wrong because k√p = sign(k)*√(k²p) So a*√(a²) = sign(a)*sqrt(a⁴)
    in does | a | x a = a^2? Comment by indica May 2021